In the chemical reaction between stoichiometic quantities of `KMnO_4` and `Kl` in weakly basic solution, what is the number of moles of `I_2` released for 4 moles of `KMnO_4` consumed?

Explanation

To solve the problem, we need to determine the number of moles of \( I_2 \) produced when 4 moles of \( KMnO_4 \) are consumed in a reaction with \( KI \) in a weakly basic solution.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between \( KMnO_4 \) and \( KI \) in a weakly basic medium can be represented as:
\[ 2 KMnO_4 + 6 KI \rightarrow 2 MnO_2 + 3 I_2 + 6 KOH \]
This equation shows that 2 moles of \( KMnO_4 \) react with 6 moles of \( KI \) to produce 3 moles of \( I_2 \).

2. **Determine the Stoichiometric Ratio:**
From the balanced equation, we can see that:
- 2 moles of \( KMnO_4 \) produce 3 moles of \( I_2 \).
- Therefore, the ratio of \( KMnO_4 \) to \( I_2 \) is \( \frac{3}{2} \).

3. **Calculate the Moles of \( I_2 \) from 4 Moles of \( KMnO_4 \):**
If we have 4 moles of \( KMnO_4 \), we can use the stoichiometric ratio to find the moles of \( I_2 \) produced:
\[ \text{Moles of } I_2 = 4 \text{ moles of } KMnO_4 \times \frac{3 \text{ moles of } I_2}{2 \text{ moles of } KMnO_4} \]
\[ = 4 \times \frac{3}{2} = 6 \text{ moles of } I_2 \]

4. **Final Answer:**
Therefore, when 4 moles of \( KMnO_4 \) are consumed, 6 moles of \( I_2 \) are released.

### Summary:
The number of moles of \( I_2 \) released for 4 moles of \( KMnO_4 \) consumed is **6 moles**.